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# Coordinate Patterns

#### Charlie's Squares

#### More Squares from Charlie

#### A recap of rules for the n^{th} term

Like some of the students featured above, Ivan Ivanov from the 47th High School in Sofia, Bulgaria put his reasoning into algebraic form. He found formulas for the coordinates of the shapes using the nice notation below. Well done to everyone who found similar formulae.

Charlie's Squares

#### Alison's Triangles

#### More Squares from Charlie

If $B(n)$ is the centre of square $nb$, then the coordinates of $B(n)$ satisfy the equations: $x(n) = 2n$, and $y(n)B = -2n + 8$.

You might like to think if you can find a formula for squares such as 19c or 199a. What about if Charlie continues the pattern and draws squares such as 1d and 1e, etc?

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Age 11 to 14

Challenge Level

- Problem
- Getting Started
- Student Solutions
- Teachers' Resources

Wow, we received loads of solutions here! Let's have a look at a few of them:

William and Chris, from Croftlands Junior School, tried to approach Charlie's Squares this way:

We found that the best way to organise our information was to draw a table. Here is the table we drew:

Square | x-coordinate | y-coordinate |

1 | 2 | 2 |

2 | 5 | 3 |

3 | 8 | 4 |

4 | 11 | 5 |

5 | 14 | 6 |

Using the table, we found out that the x-coordinate was going up 3 every time. We added a couple of extra examples in the table. We spotted that the y-coordinate was the number of the square plus 1. Then we tried spotting patterns by working with the y coordinates to find the x coordinates.

A great way to start! They then went on to find the general formula correctly. Callum, Elys, Cerys, Elgan, Cullen, Ethan, Ifan and Twm, from Ysgol Llanegryn, jumped straight in with the following:

We first found a pattern: (2,2), (5,3), (8,4), and we then discussed how to find any centre square. We turned to algebra. The n^{th} term in the pattern is (3n -1, n + 1) so the coordinates of the 20th centre point would be (60-1, 20+1) which is (59,21).

Justin, from the John of Gaunt School, correctly noted:

In the case of going to the left, the process is repeated but instead of increasing the horizontal and vertical coordinates, the coordinates decrease by the same amounts.

Sam, from Fern Avenue, gave the following answer to Alison's Triangles:

I started by comparing triangles 1 and 3, and realised that 'middle' vertices on nearby odd triangles were exactly 8 points horizontally apart. That, with the fact that all odd middle vertices have coordinates of the form (x,10), allows you to find the middle vertex of any odd numbered triangle.

The same applies for all even numbered triangles, except that the vertices are on coordinates that are 4 grid squares to the right of the odd triangles.

After noticing this, I realised that the even isosceles triangles continued to the middle vertex of the odd numbered triangles, and vice-versa. This meant that they either went down or up 5 squares, and right 2 squares.

I now realised that if I knew the coordinates of any triangle, then I knew the middle vertex of the next triangle.

Therefore, triangle 23 will have vertices at (88,5), (90,10), and (92,5).

Joe and Jack, from Springfield Primary School, gave us their formula for the x-coordinate of the top/bottom vertex of the n^{th} triangle, which was 4n-2. From this they plugged in n = 23, which gave them the correct x-coordinate, and noticed that the y-coordinate alternates between 10 (when n was odd) and 0 (when n was even). Great!

Penny, Jacob, Patrick and Cameron, from Inter Lakes, submitted solutions to More Squares from Charlie. Cameron wrote:

The coordinates of the centre of square 22b are (44,-36). To work this out, first I made a chart for the x-coordinates and the y-coordinates. To get from one 'b' square to the next, add 2 to the x-coordinate and subtract 2 from the y-coordinate. My quick and efficient strategy is knowing the x and y axis rules.

Brilliant!

Charlie's Squares

- If $C(n)$ is the centre of square $n$,

then the coordinates of $C(n)$ satisfy the equations: $x(n) = 3n - 1$, and $y(n) = n + 1$. - If $L(n)$ is the bottom left hand vertex of square $n$,

then the coordinates of $L(n)$ satisfy the equations: $x(n) = 3n - 2$, and $y(n) = n - 1$.

- If $C(n)$ is the top or bottom vertex of triangle $n$ (for odd and even $n$ respectively), then the coordinates of $C(n)$ satisfy the equations: $x(n) = 4n - 2$, $y(n) = 10$ when $n$ is odd and $y(n) = 0$ when $n$ is even.
- If $L(n)$ is the left-most vertex of triangle $n$, then the coordinates of $L(n)$ satisfy the equations: $x(n) = 4n - 4$, $y(n) = 5$.
- The right-most vertex of triangle $n$ is the same as the left-most vertex of triangle $n-1$.

If $B(n)$ is the centre of square $nb$, then the coordinates of $B(n)$ satisfy the equations: $x(n) = 2n$, and $y(n)B = -2n + 8$.

You might like to think if you can find a formula for squares such as 19c or 199a. What about if Charlie continues the pattern and draws squares such as 1d and 1e, etc?

Michael Sena from NSBH sent in a little computer program that could give the coordinates of any of Charlie's first set of squares, even producing a table of the first few. Well done!

Thanks again to everyone for their solutions!

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